Why isn't the snub tesseract uniform?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Why isn't the snub tesseract uniform?

Postby ubersketch » Sun Dec 24, 2017 5:40 pm

Read the title.
Same for all the other snub polychora.
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Re: Why isn't the snub tesseract uniform?

Postby Mercurial, the Spectre » Sun Dec 24, 2017 8:23 pm

When you snub a generic polychoron/honeycomb of the form {p,q,r}, that is, alternating is omnitruncation, you will have 6 different edge lengths to work out on. You have only four degrees of freedom between variations, so the other two are neglected. That is, they cannot have equal edges in general.

For the s3s3s4s (snub tesseract), you have to consider the topology of the x3x3x4x (omnitruncated tesseract).
The x3x3x4x has:
x3x3x - truncated octahedron (Td symmetry)
x3x2x - hexagonal prism (D3h symmetry)
x2x4x - octagonal prism (D4h symmetry)
x3x4x - truncated cuboctahedron (Oh symmetry)
while the s3s3s4s has:
s3s3s - icosahedron (T symmetry)
s3s2s - octahedron (D3 symmetry)
s2s4s - square antiprism (D4 symmetry)
s3s4s - snub cube (O symmetry)

To rigorously prove that s3s3s4s cannot be made uniform, look at the truncated cuboctahedron of which the uniform snub cube is based (look at Wikipedia). The vertices of the related x3x4x are simply the union of the snub cube's enantiomorphs. The x4x (octagon) in the unique x3x4x, from which its alternation is the uniform snub cube, is not a regular octagon. Rather, it is merely an octagon with equal angles and 2 different kinds of edges that alternate with each other with a symmetry equal to a square.

Now look at the uniform square antiprism. It is derived from x2x4x, except that x4x is always a regular octagon (since a uniform antiprism's two bases must be opposite with respect to each other in a dual fashion). In the omnitruncated tesseract, you have the fact that every x2x4x is connected to x3x4x through an octagon. Since individually, the octagons needed in making them uniform are different but dependent on one in the omnitruncate itself, it cannot be made uniform because it only takes on one value, not two.

The only uniform cases therefore belong to s2s2s2s (hex), s3s3s *b3s (sadi) and s5s2s(5/3)s (gudap), because the base polytope is regular (meaning equal faces) or because there exists a higher symmetry in which the number of edge lengths is the same or lesser than the degrees of freedom, from which a solution can be easily determined.
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Re: Why isn't the snub tesseract uniform?

Postby Klitzing » Sun Dec 24, 2017 10:14 pm

In fact you have to understand that snubbing for usual in fact is an application of 2 processes. The one is the alternated faceting, the other is a topological variation of the edge sizes. In fact you could do it both ways: Either apply the variation first (onto the omnitruncate) and thereafter apply the alternated faceting, which then ought result directly in the uniform snub, or you could do it the other way round, i.e. first apply the alternation and thereafter do the variation of the outcome.

You already understood - or at least Mercurial explained it to you - that the variation process depends highly on the actual degree of freedom. Thus in the older literature often it is said, that snubbing is not applicable at all in most cases of 4D and beyond. This is because of the assumption of the first mentioned order of applications. There the initial variation process cannot be applied because of missing degrees of freedom. But, when using the second mentioned order of applications, you see that the alternating faceting is applicable in any case. Thus the topology of the snub already can be obtained. It then is only the thereafter to be applied variation back to uniformity, which has to bow under the existing degrees of freedom.

So it is well allowed to consider a figure like s3s3s4s in the sense of a mere alternated faceting. But sadly in that very case it happens that s3s3s4s cannot be made uniform, as Mercurial already stated.

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